XAT 2025 Number System Questions

XAT 2025 number system questions will aid students in learning more about the important topics in the XAT 2025 syllabus. Students will be able to analyse and understand common questions that have been appearing for years by solving the questions. Some of the important topics in the XAT number system are Algebra, Number System, LCM & HCF, Profit & Loss, Geometry and so on. There will be 28 questions coming from Quantitative Ability & Data Interpretation. In this article know about the XAT 2025 top number system questions stated below.

XAT Number System Questions

XAT number system questions will help students understand the paper's difficulty level and the important chapters to study. Students can go through top 25 CAT number system questions given below.

Question 1: What is the difference of median and mean of the given data: 4, 8, 13, 15, 9, 21, 18, 23, 1, 35?

a) 0.7

b) 1.7

c) 1.2

d) 2.1

1) Answer (A)

Solution:

Mean:

No. of samples (n) = 10

Mean = ∑xn=4+13+8+15+9+21+18+23+35+110=14710=14.7

Median:

By arranging the in ascending order, we get:

1, 4, 8, 9, 13, 15, 18, 21, 23, 35

n = 10 (even)

Therefore, the median is the average of the 5th and 6th terms.

Median = 13+152=14

Mean – Median = 14.7 – 14 = 0.7

Therefore, Option A is correct.

Question 2: If 60% of a number is 168, then tellt is the number?

a) 280

b) 320

c) 240

d) 200

2) Answer (A)

Solution:

60% of the number is 168.

Let’s imagine the number is ‘y’.

60% of y = 168

0.6y = 168

y = 280

Question 3: Calculate the value of 5÷5+5–6÷3×4+2+(3÷6×2)?

a) 21

b) 25

c) 28

d) 19

3) Answer (B)

Solution:

5×5×55+5−63×4+2+36×2

25+5−8+2+1

25

Question 4: Calculate the median of the given data?
41, 43, 46, 50, 85, 61, 76, 55, 68, 95

a) 61

b) 58

c) 57

d) 55

4) Answer (B)

Solution:

Rearrange the given data in ascending order.

41, 43, 46, 50, 55, 61, 68, 76, 85, 95

n = number of given data

n = 10

median = (n2)th term + (n2+1)th term2

= (102)th term + (102+1)th term2

= 5th term + (5+1)th term2

= 5th term + 6th term2

= 55+612

= 1162

= 58

Question 5: What is the value of (B – A) if A is the lowest three-digit number that is divisible by both 6 and 7 and B is the largest four-digit number that is divisible by both 6 and 7?

a) 9912

b) 9870

c) 9996

d) 9954

5) Answer (B)

Solution:

LCM of (6, 7) = 42

6 = 2×3

7 = 7

Based on the facts in the question, A and B's values can be divided by both 6 and 7. These have an LCM of 42. This indicates that A and B's values will both be multiples of 42.

Suppose that A is the lowest three-digit number that divides by 6 and 7.

The smallest three-digit number is 100. When we divide 100 by 42, then the remainder will be 16. So we need to add (42-16) in 100.

So A = 100+(42-16)

A = 100+26

A = 126

B is the largest four-digit number divisible by both 6 and 7.

The largest four-digit number is 9999. When we 9999 by 42, then the remainder will be 3. So we need to subtract 3 from 9999.

So B = 9999-3

B = 9996

Value of (B – A) = 9996-126

= 9870

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Question 6: The following data set is represented as follows: 4, 2, 3, 2, 7, 4, 8, 5, 2, 4, 5, 6, 2, 5, 6, 6, 5, 4, 6, 5, 3, 5, 4, 3 What is the set's mode?

a) 2

b) 5

c) 6

d) 4

6) Answer (B)

Solution:

4, 2, 3, 2, 7, 4, 8, 5, 2, 4, 5, 6, 2, 5, 6, 6, 5, 4, 6, 5, 3, 5, 4, 3

Arrange the given data in ascending order.

2, 2, 2, 2, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7, 8

Mode = maximum number of times a number is available in the given data set

= 5 (It is available six times in the given data set.)

Question 7:Calculate the first seven prime numbers' median is:

a) 7

b) 5

c) 13

d) 11

7) Answer (A)

Solution:

first, the seven prime numbers are 2, 3, 5, 7, 11, 13, 17.

Here the number of terms is 7. So n = 7.

median = (n+12)th term

= (7+12)th term

= (82)th term

= 4th term

= 7

Question 8: What is the largest six-digit number that can be divided by three, four, five, and six times its sum of digits?

a) 45

b) 42

c) 39

d) 48

8) Answer (B)

Solution:

As we know the largest six-digit number is 999999.

The required number should be divisible by 3, 4, 5 and 6.

So the LCM of 3, 4, 5 and 6 is 60.

When 999999 is divided by 60, then the remainder will be 39.

999999=60×16666+39

Required number = 999999-39 = 999960

the sum of the digits of the required number = 9+9+9+9+6+0 = 42

Question 9: What is the difference between the mean and median of the given data:
4, 6, 3, 7, 10, 13, 16 and 5?

a) 5

b) 1.5

c) 3

d) 4.5

Answer (B)

9. What is the difference between the mean and median of the given data: 4, 6, 3, 7, 10, 13, 16 and 5?

1. 5
2. 1.5
3. 3
4. 4.5

Solution:

Mean=sum of data number of data

=4+6+3+7+10+13+16+58

=648

= 8

For the median, first, arrange the given data in ascending order from left to right.

3, 4, 5, 6, 7, 10, 13, 16

n = number of data = 8

median = (n2)th term + (n2+1)th term2

= (82)th term + (82+1)th term2

= 4th term + (4+1)th term2

= 4th term + 5th term2

= 6+72

= 132

= 6.5

difference between the mean and median = 8-6.5

= 1.5

Question 10: What distinguishes the following data sets' means from medians: 5, 7, 8, 13, 12, 14, 9, 2, 26, 10?

a) 2.3

b) 0.4

c) 1.8

d) 1.1

10) Answer (D)

Solution:

First, arrange the given data in ascending order from left to right.

2, 5, 7, 8, 9, 10, 12, 13, 14, 26

There are ten numbers. So n = 10.

median = (n2)th term + (n2+1)th term2

= (102)th term + (102+1)th term2

= 5th term + (5+1)th term2

= 5th term + 6th term2

= 9+102

= 192

= 9.5

mean = sum of data number of data

= 2+5+7+8+9+10+12+13+14+2610

= 10610

= 10.6

difference between the mean and the median = 10.6-9.5

= 1.1

Question 11: If x be the median of data: 33, 42, 28, 49, 32, 37, 52, 57, 35, 41.
If 32 is replaced by 36 and 41 by 63, then the median of the data, so obtained, is y. What is the value of (x + y)?

a) 78

b) 78.5

c) 79.5

d) 79

11) Answer (B)

Solution:

33, 42, 28, 49, 32, 37, 52, 57, 35, 41

Arrange the given data in ascending order from left to right.

28, 32, 33, 35, 37, 41, 42, 49, 52, 57

Median = (n2)th term +(n2+1)th term2

Here n = the number of data

Median = x = (102)th term +(102+1)th term2

= 5th term +(5+1)th term2

= 5th term +6th term2

= 37+412

= 782

x = 39

If 32 is replaced by 36 and 41 by 63, then the median of the data, so obtained, is y.

28, 36, 33, 35, 37, 63, 42, 49, 52, 57

Arrange the given data in ascending order from left to right.

28, 33, 35, 36, 37, 42, 49, 52, 57, 63

Median = y = (102)th term +(102+1)th term2

= 5th term +(5+1)th term2

= 5th term +6th term2

= 37+422

= 792

y = 39.5

value of (x + y) = (39+39.5)

= 78.5

Question 12: The following table shows the weight of 20 students:

The mode and median of the above data are, respectively:

a) 51 and 48

b) 53 and 56

c) 60 and 53

d) 48 and 53

Answer (D)

12) The following table shows the weight of 20 students:

Solution:

Mode is the maximum number of students who have the same weight. So here 6 students have 48 kg weight. Hence 48 will be the mode.

For Median, first, we need to arrange the given weight in the ascending order from left to right.

Total number of students = n = 20

Median = (n2)th+(n2+1)th2

= (202)th+(202+1)th2

= (10)th+(10+1)th2

= (10)th+(11)th2

= 53+532

= 53

Question 13: The mode of the given data 2, 5, 5, 7, 2, 6, 8, 6, 9, 6 is:

a) 7

b) 6

c) 5

d) 2

Answer (B)

13) The supplied data set (2, 5, 5, 7, 2, 6, 8, 6, 9, 6) has the following mode:

1. 7
2. 6
3. 5
4. 2

Solution:

Arrange the given data in ascending order which is given below.

2, 2, 5, 5, 6, 6, 6, 7, 8, 9

The maximum number of times a number is occurring is called mode. Here ‘6’ is occurring three times. So ‘6’ will be the mode.

Question 14: x is the greatest number by which, when 2460, 2633 and 2806 are divided, the remainder in each case is the same. What is the sum of digits of x?

a) 11

b) 10

c) 13

d) 9

14) Answer (A)

Solution:

As given in the question, the remainder is the same in each. So (2633-2460), (2806-2460) and (2806-2633) should be completely divisible by ‘x’.

(2633-2460) = 173

(2806-2460) = 346

(2806-2633) = 173

As we know 173 is a prime number. So the HCF of (173, 346 and 173) will be 173.

Hence x = 173

The sum of digits of x = 1+7+3

= 11

Question 15: Find the sum of the median and mode of the data
8, 1, 5, 4, 9, 6, 3, 6, 1, 3, 6, 9, 1, 7, 2, 6, 5?

a) 13

b) 11

c) 12

d) 14

15) Answer (B)

Solution:

Arrange the given data in ascending order which is given below.

1, 1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 6, 6, 7, 8, 9, 9

Median = middle term of the given data after arranging them in ascending order

Here 17 numbers are given. So the median will be the 9th number from the left end. It’s 5.

Mode = maximum number of times a number is occurring in the given data

= 6 (It is coming four times in the given data)

The sum of the median and mode of the data = 5+6

= 11

Question 16: Find the mode of 2, 2, 3, 3, 5, 5, 5, 7, 8, 8, 9, 10 is:

a) 5

b) 2

c) 3

d) 6

16) Answer (A)

Solution:

Mode: The value that appears most often in a set of given data values.

Given Data: 2, 2, 3, 3, 5, 5, 5, 7, 8, 8, 9, 10

• The most number repeated in the above data is 5.

So, the Mode of the given data is 5.

Hence, Option A is correct.

Question 17: The mode of the following data is 36. What is the value of x?

a) 11

b) 15

c) 13

d) 12

Answer (D)

17) The mode of the following data is 36. What is the value of x ?

Solution:

As per given data,

The class interval of 30-40 has the highest frequency, that it is a modal class

As we know,

M = l+{(f1−f0)2f1−f0−f2}× h

where, h= size of the class interval,

l = lower limit of the modal class,

f1= frequency of the modal class,

f0= frequency of the class preceding the modal class

f2= frequency of the class succeeding the modal class

putting the values from the given data :

36=30+(16−10)2× 16−10−x× 10

36−30=622−x× 10

22−x=10

x=12

Hence, Option D is correct.

Question 18: The remainder in each case of 6892, 7105, and 7531 divided by the largest number x equals y. How much does (x – y) mean?

a) 123

b) 137

c) 147

d) 113

18) Answer (B)

Solution:

We have to find HCF of given numbers: 6892, 7105, 7531

7105 – 6892 = 213

7531 – 7105 = 426

426 – 213 = 213

So, Either the difference or the factor of difference is the HCF of those given numbers.

Here, 213 is the HCF.

When 6892, 7105, and 7531 is divided by 213 we get 76 as a remainder

So, x = 213 and y = 76

According to Question :

x – y = 213 – 76 = 137

Hence, Option B is correct. 

Question 19: The perfect square between 120 and 300 adds up to:

a) 1400

b) 1024

c) 1296

d) 1204

19) Answer (A)

Solution:

The sum of the squares of n consecutive numbers =

The sum of the perfect square between 120 and 300 = 112+122+132+142+152+162+172

=17(17+1)(2(17+1))6−10(10+1)(2(10)+1)6

=17(18)(35)6−10(11)(21)6

=51×35−11×35

=35(51−11)

=35(40)

=1400

Hence, the correct answer is Option A

Question 20: The following is the difference between the largest and least four-digit numbers that start with 3 and finish with 5:

a) 990

b) 900

c) 909

d) 999

20) Answer (A)

Solution:

The greatest four-digit number that begins with 3 and ends with 5 = 3995

The least four-digit number that begins with 3 and ends with 5 = 3005

∴ The difference between the greatest and the least four-digit numbers that begin with 3 and end with 5 = 3995 – 3005 = 990

Question 21: If a nine-digit number 389 x 6378 y is divisible by 72, then what is the value of 6x+7y:

a) 6

b) 13

c) 46

d) 8

Answer (D)

21) If a nine-digit number 389 x 6378 y is divisible by 72, then the value of 6

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